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Friday, 07 July 2017 03:56

Swap nodes in a linked list without swapping data

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Given a linked list and two keys in it, swap nodes for two given keys. Nodes should be swapped by changing links. Swapping data of nodes may be expensive in many situations when data contains many fields.

It may be assumed that all keys in linked list are distinct.


Input:  10->15->12->13->20->14,  x = 12, y = 20
Output: 10->15->20->13->12->14

Input:  10->15->12->13->20->14,  x = 10, y = 20
Output: 20->15->12->13->10->14

Input:  10->15->12->13->20->14,  x = 12, y = 13
Output: 10->15->13->12->20->14

This may look a simple problem, but is interesting question as it has following cases to be handled.
1) x and y may or may not be adjacent.
2) Either x or y may be a head node.
3) Either x or y may be last node.
4) x and/or y may not be present in linked list.

How to write a clean working code that handles all of the above possibilities.

We strongly recommend to minimize your browser and try this yourself first.

The idea it to first search x and y in given linked list. If any of them is not present, then return. While searching for x and y, keep track of current and previous pointers. First change next of previous pointers, then change next of current pointers. Following are C and Java implementations of this approach.

/* This program swaps the nodes of linked list rather
   than swapping the field from the nodes.*/
/* A linked list node */
struct Node
    int data;
    struct Node *next;
/* Function to swap nodes x and y in linked list by
   changing links */
void swapNodes(struct Node **head_ref, int x, int y)
   // Nothing to do if x and y are same
   if (x == y) return;
   // Search for x (keep track of prevX and CurrX
   struct Node *prevX = NULL, *currX = *head_ref;
   while (currX && currX->data != x)
       prevX = currX;
       currX = currX->next;
   // Search for y (keep track of prevY and CurrY
   struct Node *prevY = NULL, *currY = *head_ref;
   while (currY && currY->data != y)
       prevY = currY;
       currY = currY->next;
   // If either x or y is not present, nothing to do
   if (currX == NULL || currY == NULL)
   // If x is not head of linked list
   if (prevX != NULL)
       prevX->next = currY;
   else // Else make y as new head
       *head_ref = currY; 
   // If y is not head of linked list
   if (prevY != NULL)
       prevY->next = currX;
   else  // Else make x as new head
       *head_ref = currX;
   // Swap next pointers
   struct Node *temp = currY->next;
   currY->next = currX->next;
   currX->next  = temp;
/* Function to add a node at the begining of List */
void push(struct Node** head_ref, int new_data)
    /* allocate node */
    struct Node* new_node =
        (struct Node*) malloc(sizeof(struct Node));
    /* put in the data  */
    new_node->data  = new_data;
    /* link the old list off the new node */
    new_node->next = (*head_ref);
    /* move the head to point to the new node */
    (*head_ref)    = new_node;
/* Function to print nodes in a given linked list */
void printList(struct Node *node)
    while(node != NULL)
        printf("%d ", node->data);
        node = node->next;
/* Druver program to test above function */
int main()
    struct Node *start = NULL;
    /* The constructed linked list is:
     1->2->3->4->5->6->7 */
    push(&start, 7);
    push(&start, 6);
    push(&start, 5);
    push(&start, 4);
    push(&start, 3);
    push(&start, 2);
    push(&start, 1);
    printf("\n Linked list before calling swapNodes() ");
    swapNodes(&start, 4, 3);
    printf("\n Linked list after calling swapNodes() ");
    return 0;
 Linked list before calling swapNodes() 1 2 3 4 5 6 7
 Linked list after calling swapNodes() 1 2 4 3 5 6 7

Optimizations: The above code can be optimized to search x and y in single traversal. Two loops are used to keep program simple.

Read 1563 times Last modified on Friday, 07 July 2017 04:04
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